//给定一个整数数组 A，返回其中元素之和可被 K 整除的（连续、非空）子数组的数目。 
//
// 
//
// 示例： 
//
// 输入：A = [4,5,0,-2,-3,1], K = 5
//输出：7
//解释：
//有 7 个子数组满足其元素之和可被 K = 5 整除：
//[4, 5, 0, -2, -3, 1], [5], [5, 0], [5, 0, -2, -3], [0], [0, -2, -3], [-2, -3]
// 
//
// 
//
// 提示： 
//
// 
// 1 <= A.length <= 30000 
// -10000 <= A[i] <= 10000 
// 2 <= K <= 10000 
// 
// Related Topics 数组 哈希表 前缀和 
// 👍 302 👎 0

/**
 * @author DaHuangXiao
 */
package leetcode.editor.cn;

import java.util.HashMap;
import java.util.Map;

public class SubarraySumsDivisibleByK {
    public static void main(String[] args) {
        Solution solution = new SubarraySumsDivisibleByK().new Solution();
        System.out.println(solution.subarraysDivByK(new int[]{4, 5, 0, -2, -3, 1}, 5));
    }
    //leetcode submit region begin(Prohibit modification and deletion)
class Solution {
    public int subarraysDivByK(int[] nums, int k) {
        Map<Integer,Integer> map = new HashMap<>();
        map.put(0,1);
        int total = 0;
        int res = 0;
        for (int i = 0; i < nums.length; i++) {
            total = total+nums[i];


            int c = 0;

            while (c*k<=total) {
                res+=map.getOrDefault(total-c*k,0);
                res+=map.getOrDefault(total+c*k,0);
                c++;
            }
            map.put(total,map.getOrDefault(total,0)+1);
        }
        return res;
    }
}
//leetcode submit region end(Prohibit modification and deletion)

}